If a^⅓+b^⅓+c^⅓=0 , prove that (a+b+c)³=27abc
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Solution:
It is given that,
a^⅓+b^⅓+c^⅓=0
=> a^⅓+b^⅓= -c^⅓---(1)
=> (a^⅓+b^⅓)³=(-c^⅓)³
=> (a^⅓)³+(b^⅓)³+3a^⅓b^⅓[a^⅓+b^⅓] =-c
=> a+b+3a^⅓b^⅓(-c^⅓)= -c
/* from (1) */
=> a+b+c = 3a^⅓b^⅓c^⅓
=> (a+b+c)³ = (3a^⅓b^⅓c^⅓)³
=> (a+b+c)³ = 27abc
Hence , prooved
••••
It is given that,
a^⅓+b^⅓+c^⅓=0
=> a^⅓+b^⅓= -c^⅓---(1)
=> (a^⅓+b^⅓)³=(-c^⅓)³
=> (a^⅓)³+(b^⅓)³+3a^⅓b^⅓[a^⅓+b^⅓] =-c
=> a+b+3a^⅓b^⅓(-c^⅓)= -c
/* from (1) */
=> a+b+c = 3a^⅓b^⅓c^⅓
=> (a+b+c)³ = (3a^⅓b^⅓c^⅓)³
=> (a+b+c)³ = 27abc
Hence , prooved
••••
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brunoconti:
did u have more than 3 questions ?
no
only had 3
and thanks
thk you
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